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ganeshie8
 one year ago
consider a regular \(ngon\).
show that a "flip", followed by a "ccw rotation" equals the "cw rotation" followed by a "flip":
\[\large rf=fr^{1}\]
ganeshie8
 one year ago
consider a regular \(ngon\). show that a "flip", followed by a "ccw rotation" equals the "cw rotation" followed by a "flip": \[\large rf=fr^{1}\]

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perl
 one year ago
Best ResponseYou've already chosen the best response.4That's a concrete example, to prove it abstractly ...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440623844373:dw

perl
 one year ago
Best ResponseYou've already chosen the best response.4so we want to show r * f = f * r^1 , where * stands for composition we can use the fact that f*f = Identity

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i think so :) \(f = f^{1}\) is kinda obvious as reflection undoes itself..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://en.wikipedia.org/wiki/Dihedral_group#Group_structure

perl
 one year ago
Best ResponseYou've already chosen the best response.4r * f = r *f^1 = (r^1)^1 * f^1 = ( f * r^1 )^1

perl
 one year ago
Best ResponseYou've already chosen the best response.4Apparently I proved that r*f and f*r^1 are inverses, but that wasn't the question :)

perl
 one year ago
Best ResponseYou've already chosen the best response.4using (a*b)^1 = b^1 * a^1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Haha nice :) that is also clear from shoe socks principle \((rf)^{1} = f^{1}r^{1} = fr^{1}\)

perl
 one year ago
Best ResponseYou've already chosen the best response.4is there a proof r f = f r^1 without having to go into the geometry , considering when n is even and odd, etc

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1not so sure, i got stuck on this while attempting below exercises https://i.gyazo.com/6b1fbd9dc197477edffe9987ca355527.png

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1they are from Gallian abstract algebra book

perl
 one year ago
Best ResponseYou've already chosen the best response.4You can use the fact that frf = r^1

perl
 one year ago
Best ResponseYou've already chosen the best response.4given f*f = I (f * r* f) = r^1 r* f = I * (r*f) = (f* f )* r * f = f * ( f* r * f ) = f * r ^1

perl
 one year ago
Best ResponseYou've already chosen the best response.4'given' meaning these are theorems for rigid transformations of ngon

perl
 one year ago
Best ResponseYou've already chosen the best response.4f r f = r^1 In geometric terms: in the mirror a rotation looks like an inverse rotation.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that looks interesting!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1but aren't we trying to prove the same thing

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\large rf=fr^{1}\] left multiplying both sides by \(f\) gives \[\large frf=r^{1}\]